Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. A car on a curved exit ramp
Read the story as text
A car takes a highway exit ramp. The ramp is a curve with a radius of 50 m, and the driver keeps a steady 15 m/s.
The speedometer never moves. Yet the passengers feel pressed against the door, and on a rainy night the same car,
at the same speed, slides off toward the outside of the curve.
The speed is constant, so why do we need a force at all? Which force is it? And why does the rain matter?
What makes an object move in a circle, and how big must that push or pull be?
2. A ball whirled on a string
3. Over the top of a hill
Quick check. Seen from above, Ana's ball whirls around the peg. The string snaps just as the ball moves straight to the right. Which way does the ball go?
Another example: Letting go of the spinning ball
Another example: Pressed against the wall of a turning bus
4. Twice as fast, four times the pull
5. One trip around the merry-go-round
6. Upside down at the top of a loop
Quick check. Leo's ball goes around the same circle, now three times as fast. How does the string's pull change?
Another example: Twice as fast around the bike path curve
7. Check yourself
Think of your answer first, then tap to see it.
a) A ball on a string moves in a circle at a constant speed. Is it accelerating?
Show answer
Yes. Its direction keeps changing, so its velocity changes. The acceleration points to the center.
b) The string breaks. Which way does the ball go?
Show answer
In a straight line along its velocity at that moment, tangent to the circle, not straight outward.
c) A 0.5 kg ball moves at 4 m/s in a circle of radius 1.0 m. How big is the net force toward the center?
Show answer
F = m v² ÷ r = 0.5 kg × (4 m/s)² ÷ 1.0 m = 8 N, toward the center.
Already know this?
Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.
1. A car drives around a curve at a constant 20 m/s. Is it accelerating?
2. A car turns on a flat road. Which real force points toward the center and keeps it on the curve?
3. A 0.40 kg ball moves at 3.0 m/s on a string 0.60 m long, in a horizontal circle on a smooth table. How hard does the string pull?
The idea in plain words
Velocity is a vector. In a circle the speed can stay the same while the direction keeps changing.
A changing velocity means there is an acceleration. For an object moving at constant speed v on a circle of radius r
(uniform circular motion), the acceleration points toward the center and has size
ac = v² / r (toward the center, "centripetal" = center-seeking)
One trip around takes the period T. In that time the object covers the circumference 2πr, so
v = 2πr / T
Newton's second law still runs the show. To get an acceleration toward the center, the net force must point
toward the center:
Fnet, toward center = m v² / r
This is not a new kind of force. It is the job that real forces do. Ask: which real force (or which mix of forces)
points toward the center?
Trap: drawing an extra "centripetal force" arrow on the free body diagram. Instead: draw only real forces (tension, friction, normal, gravity); the part of their sum toward the center is mv²/r.
Ball on a string: the tension.
Car on a flat curve: static friction from the road, pointing toward the center. Max friction is μs m g, so the fastest safe speed is vmax = √(μs g r). Rain lowers μs, so it lowers vmax.
Top of a loop-the-loop: both the normal force and gravity point down, toward the center: N + m g = m v²/r. The slowest the cart can go and still touch the track is when N = 0: vmin = √(g r).
Banked curve: the road is tilted, so the normal force leans toward the center. At the design speed no friction is needed: tan θ = v² / (r g).
Satellite: gravity alone. G M m / r² = m v² / r, so v = √(G M / r).
Trap: drawing an outward "centrifugal force" because you feel thrown outward in a turning car. Instead: nothing pushes you outward; your body tends to go straight (first law) while the door or seat pushes you inward.
If the inward force disappears (the string breaks, the road turns to ice), the object does not fly straight out from the center.
It keeps the velocity it had at that moment and moves in a straight line along the tangent (Newton's first law).
Trap: saying a released ball flies straight outward, away from the center. Instead: it leaves along the tangent, in the direction it was moving at that instant.
Worked with numbers: the exit ramp, dry and wet
Read the steps as text
Acceleration toward the center: ac = v²/r = (15 m/s)² / 50 m = 225/50 = 4.5 m/s².
Constant speed, but the direction turns, so there is an acceleration, pointing at the center of the curve.
Net inward force on the 1200 kg car: F = m ac = 1200 × 4.5 = 5400 N.
Second law, applied along the direction toward the center.
Which real force? On a flat road the only horizontal force is static friction from the road on the tires. So fs = 5400 N toward the center.
Weight (down) and normal force (up) are vertical and cancel: N = m g = 11 760 N.
Dry road, μs = 0.80: max friction = 0.80 × 11 760 = 9408 N. Since 5400 N ≤ 9408 N, the car holds the curve.
Static friction can be anything from 0 up to μs N. It only supplies what is needed.
Fastest safe speed when dry: μs m g = m v²/r → vmax = √(μs g r) = √(0.80 × 9.8 × 50) = √392 = 19.8 m/s.
The mass cancels: a truck and a small car have the same top speed on the same curve (same tires and road).
Wet road, μs = 0.40: vmax = √(0.40 × 9.8 × 50) = √196 = 14 m/s. At 15 m/s the car needs 5400 N but friction can give only 0.40 × 11 760 = 4704 N, so it slides outward.
Not enough inward force means the path bends less than the curve: the car drifts to the outside. Try it in animation 2.
Try it: animations
Predict first, then press Play. The free body diagram next to each scene shows only real forces. There is never a separate "centripetal force" arrow: the net of the real forces points to the center.
1. A puck on a string (top view, frictionless air table)
The string holds the puck on a circle. Choose when the string breaks and watch the path afterward. Weight and the table's upward push cancel; they point into and out of the screen, so the diagram shows only the tension.
Free body diagram (top view)
Tension = m v²/r, always toward the center. Double the speed and the tension becomes 4 times bigger. After the break the diagram is empty: no horizontal force, so constant velocity.
2. A car on a curve (top view, with a bank angle)
Lower μs to 0.4 for a wet road, or raise the speed, and see when friction is not enough. Then add a bank angle: the tilted road lets the normal force help. The diagram is a cross-section; toward the center of the curve is to the right in the diagram.
Free body diagram (cross-section of the car)
When the car slides, the model uses kinetic friction with μk = 0.75 μs. With a bank, at the design speed √(r g tanθ) the friction arrow disappears.
3. Loop-the-loop
A cart enters a vertical loop at the bottom (friction ignored). Is it fast enough to stay on the track at the top? The default 22.2 m/s is just above √(5 g R) = 22.1 m/s for R = 10 m. Try 18 m/s and 12 m/s.
Free body diagram of the cart
At the bottom N is bigger than m g (riders feel heavy). At the top N and m g both point down. If N would have to be negative, the cart leaves the track and becomes a projectile.
4. A satellite launched sideways
A satellite is given a sideways speed at some height. Here 1 second of animation = 1 minute of real time (the timeline counts minutes). Gravity is the only force.
Free body diagram of the satellite
The only force is gravity toward Earth's center. At exactly v = √(GM/r) the satellite falls around the Earth on a circle; it is always falling, never landing.
Examples
Example 1: swinging a ball on a string basic
A 0.15 kg ball on a 0.80 m string goes around a horizontal circle (ignore gravity's small effect on the string) at 2.0 revolutions per second. Find (a) the period, (b) the speed, (c) the acceleration, (d) the tension.
Show solutionRead the steps as text
(a) T = 1 / (2.0 rev/s) = 0.50 s. Period is time per revolution.
(b) v = 2πr/T = 2π(0.80)/0.50 = 5.03/0.50 = 10.1 m/s. One circumference per period.
(c) ac = v²/r = (10.05)²/0.80 = 101/0.80 = 126 m/s² toward the center. About 13 g. That is why a fast-swung ball pulls hard.
(d) Tension = m ac = 0.15 × 126 = 19 N. Tension is the only horizontal force, so it is the net inward force.
Example 2: top speed on a flat curve medium
A flat curve has radius 80 m. The coefficient of static friction between tires and road is 0.60. What is the fastest a car can take the curve? Does the answer depend on the car's mass?
Show solutionRead the steps as text
Forces: weight m g down, normal N up, static friction fs toward the center. Vertical: N = m g. No vertical acceleration.
Toward the center: fs = m v²/r, and fs ≤ μs m g. At top speed fs = μs m g. The limit is when friction is at its maximum.
μs m g = m v²/r → v = √(μs g r) = √(0.60 × 9.8 × 80) = √470 = 21.7 m/s (about 78 km/h). m cancels on both sides.
No, it does not depend on mass: a heavier car needs more friction but also has a bigger normal force, by the same factor.
Example 3: top of the loop AP
A 500 kg roller-coaster car goes over the top of a vertical loop of radius 10 m, upside down, at 14 m/s. (a) Find the force from the track on the car. (b) What is the slowest speed at the top for which the car still touches the track?
Show solutionRead the steps as text
At the top the center is below the car. Take down (toward the center) as positive. Forces: weight m g down, normal N down (the track is above the car and pushes it down). The track can only push, never pull.
N + m g = m v²/r → N = m v²/r − m g = 500(14²)/10 − 500(9.8) = 9800 − 4900 = 4900 N, downward. The car needs 9800 N toward the center; gravity gives 4900 N, so the track gives the rest.
(b) Slowest speed: N = 0, so m g = m v²/r → v = √(g r) = √(9.8 × 10) = √98 = 9.9 m/s. Any slower and gravity alone is more than enough inward force: the car falls inside the circle (leaves the track).
Trap: putting the diameter (the loop is 20 m tall) into v²/r. Instead: r is the radius, measured from the center: here r = 10 m.
Example 4: a banked curve AP
A curve of radius 50 m is banked at 20°. At what speed can a car take it with no friction at all (for example, on ice)?
Show solutionRead the steps as text
With no friction, the only forces are weight m g (down) and the normal force N, which is perpendicular to the tilted road, so it leans toward the center. Tilting the road tilts N.
Vertical: N cos θ = m g. Toward the center (horizontal): N sin θ = m v²/r. The car moves on a horizontal circle, so the vertical forces balance and the horizontal part of N is the net inward force.
Divide: tan θ = v²/(r g) → v = √(r g tan θ) = √(50 × 9.8 × tan 20°) = √(50 × 9.8 × 0.364) = √178 = 13.4 m/s. Again the mass cancels. Faster than this, friction must help by pointing down the slope; slower, friction points up the slope.
Example 5: the space station AP
The space station orbits 400 km above Earth's surface. Earth's radius is 6.37×106 m and G MEarth = 3.98×1014 m³/s². Find its speed and its period.
Show solutionRead the steps as text
r = 6.37×106 + 0.40×106 = 6.77×106 m. r is measured from Earth's center.
Gravity is the net inward force: G M m/r² = m v²/r → v = √(G M/r) = √(3.98×1014 / 6.77×106) = √(5.88×107) = 7670 m/s (7.67 km/s). The satellite's mass cancels.
T = 2πr/v = 2π(6.77×106)/7670 = 5550 s = about 92 minutes. About 16 orbits per day.
Practice (AP style)
1. A car drives around a circular track at a constant speed. Which statement is true?
Show answer
The direction of the velocity keeps changing, so the velocity changes and there is an acceleration, v²/r toward the center. Constant speed only means there is no acceleration along the velocity.
2. A ball on a string moves in a horizontal circle. The speed is doubled and the radius stays the same. The tension needed becomes
Show answer
T = m v²/r. Doubling v multiplies v² by 4.
3. A puck moves counterclockwise in a circle on a frictionless table, held by a string. At the instant it is at the far right of the circle (moving straight up the page), the string breaks. Afterward the puck moves
Show answer
With no net horizontal force, velocity stays constant (first law). The velocity at the instant of the break was straight up the page, tangent to the circle. There is no outward force, so it does not move outward from the center along a radius. Staying on the circle would need an inward force.
4. A car rounds a flat, unbanked curve at constant speed without slipping. The force that points toward the center of the curve is
Show answer
The tires do not slide sideways relative to the road, so it is static friction. On a flat road the normal force is vertical. "Centripetal force" is not a separate force; here it is the static friction.
5. A roller-coaster car goes over the top of a vertical loop at exactly the minimum speed needed to stay on the track. At that instant, which forces act on the car?
Show answer
At minimum speed N = 0, so gravity alone provides m v²/r. The rider feels weightless because nothing pushes on them, but gravity is acting.
6. A stone on a string is swung in a vertical circle at nearly constant speed. Where is the string most likely to break?
Show answer
At the bottom the center is up: T − m g = m v²/r, so T = m g + m v²/r (largest). At the top T + m g = m v²/r, so T = m v²/r − m g (smallest).
7. Satellite A orbits Earth on a circle of radius r. Satellite B, with twice the mass, orbits on a circle of radius 4r. Compared with A, the speed of B is
Show answer
v = √(G M/r) does not depend on the satellite's mass. Four times the radius gives √(1/4) = 1/2 the speed.
8. A rider sits on a scale in a Ferris wheel car moving at constant speed. Compared with the rider's weight m g, the scale reading at the very bottom of the wheel is
Show answer
At the bottom the center is above the rider, so the net force must point up: N − m g = m v²/r, so N = m g + m v²/r > m g. The scale reads N.
9. (Short free response: mathematical routines) A 1500 kg car takes a flat curve of radius 40 m on a wet road where μs = 0.35.
(a) Draw (describe) the free body diagram of the car as seen from behind. (b) Find the maximum safe speed. (c) A heavier truck with the same tires takes the same curve. Is its maximum speed greater, smaller or the same? Justify without numbers.
Show answer
(a) Three forces: weight m g down, normal N up (equal in size), static friction horizontal, toward the center of the curve. No "centripetal force" arrow.
(b) μs m g = m v²/r → v = √(μs g r) = √(0.35 × 9.8 × 40) = √137 = 11.7 m/s.
(c) The same. The needed inward force m v²/r and the available friction μs m g are both proportional to mass, so mass cancels.
Scoring idea: 1 point for the correct three forces with friction toward the center and no extra forces; 1 for setting max friction equal to m v²/r; 1 for 11.7 m/s; 1 for "same" with the reason that both sides scale with m.
10. (Short free response) A bucket of water (total mass 2.0 kg) is swung in a vertical circle of radius 1.0 m. (a) What is the minimum speed at the top so the water stays in the bucket? (b) At the bottom the speed is 5.0 m/s. Find the tension in the rope there. (c) Explain why the water does not fall out at the top, even though gravity pulls it down.
Show answer
(a) At minimum speed the bucket bottom pushes on the water with zero force, so m g = m v²/r → v = √(g r) = √(9.8 × 1.0) = 3.1 m/s.
(b) At the bottom, up is toward the center: T − m g = m v²/r → T = 2.0(9.8) + 2.0(5.0²)/1.0 = 19.6 + 50 = 70 N.
(c) The water is falling: gravity accelerates it toward the center. But at the top it needs at least g of inward acceleration to follow the circle; if the bucket moves fast enough (v²/r ≥ g), the bucket's path curves down at least as fast as the water would fall on its own, so the bucket stays pressed against the water and pushes it toward the center.
Scoring idea: 1 point for N = 0 condition, 1 for 3.1 m/s, 1 for correct sign/direction in the bottom equation, 1 for 70 N, 1 for an explanation that gravity (plus the push of the bucket) is the inward force and the bucket must curve down at least as fast as the water falls.
11. (Short free response: translation between representations) Seen from above, a puck on a string moves counterclockwise at constant speed in a horizontal circle on frictionless ice. At one instant it is at the north point of the circle (the top of the picture). (a) Give the direction (north, south, east or west) of its velocity, its acceleration and the net force at that instant. (b) The string breaks at that instant. Describe the puck's path. (c) A graph of the acceleration ac against speed v for this fixed radius: is it a straight line, or a curve? What happens to ac when v doubles?
Show answer
(a) Counterclockwise from above, at the north point the puck is moving west. Acceleration: south, toward the center. Net force (the tension): south, same as the acceleration.
(b) A straight line heading west, along the tangent, at the same speed (no horizontal force acts any more). It does not move outward along the radius.
(c) A curve (an upward-opening parabola), since ac = v²/r. Doubling v makes acfour times as big.
Point guide (4 points):
1 point: velocity west.
1 point: acceleration and net force both south (toward the center).
1 point: straight-line tangent path west.
1 point: curved graph, ac times 4 when v doubles.
12. (Short free response: experimental design and analysis) To test F = mv²/r, a student whirls a 0.020 kg rubber stopper in a horizontal circle of radius 0.50 m on a string that passes through a tube. The string's other end holds hanging masses, so the tension equals their weight. She times 10 revolutions: hanging weight 0.49 N: 9.0 s; 0.98 N: 6.3 s; 1.47 N: 5.2 s. (a) Describe how to keep r fixed and get good times. (b) Calculate v and v² for each run. (c) What should be plotted to get a straight line, and what should the slope be if F = mv²/r is right? Do the data agree?
Show answer
(a) Mark the string just below the tube (a paper clip) and keep the mark just under the tube, so r stays 0.50 m. Time many turns (10 or 20) and divide, to shrink the reaction-time error; repeat each run.
(b) Period T = time / 10; v = 2πr / T. 0.49 N: T = 0.90 s, v = 3.49 m/s, v² = 12.2 m²/s². 0.98 N: T = 0.63 s, v = 4.99 m/s, v² = 24.9. 1.47 N: T = 0.52 s, v = 6.04 m/s, v² = 36.5.
(c) Plot F (vertical) against v² (horizontal). Predicted slope m/r = 0.020 / 0.50 = 0.040 kg/m. From the data: (1.47 − 0.49) / (36.5 − 12.2) = 0.98 / 24.3 = 0.040 kg/m. They agree, and the line passes close to the origin.
Point guide (4 points):
1 point: a way to keep r constant.
1 point: times many revolutions to cut timing error.
1 point: v = 2πr/T with correct values.
1 point: F vs v² with slope m/r = 0.040 kg/m and a comparison.
13. (Short free response: qualitative/quantitative translation) A car takes a flat, unbanked curve of radius r. The coefficient of static friction is μs. (a) Without numbers: if the car goes twice as fast around the same curve, how much more friction does it need? (b) Derive the top safe speed vmax in terms of μs, g and r, and explain why mass does not appear. (c) Calculate vmax for r = 40 m and μs = 0.50, and again for a curve twice as wide (r = 80 m). Does the ratio match your expression?
Show answer
(a) Friction supplies mv²/r, which grows as v²: twice the speed needs four times the friction.
(b) At the limit, μs mg = mv²/r, so vmax = √(μs g r). Mass cancels: a heavier car needs more friction, but it also presses down harder, which raises the friction limit by the same factor.
(c) r = 40 m: √(0.50 × 9.8 × 40) = √196 = 14 m/s. r = 80 m: √392 = 19.8 m/s. Ratio 19.8 / 14 = 1.41 = √2, as vmax ∝ √r predicts.
Point guide (4 points):
1 point: four times the friction.
1 point: vmax = √(μs g r).
1 point: explains why mass cancels.
1 point: 14 m/s and 19.8 m/s with the √2 check.
Common mistakes
Drawing a "centripetal force" on the free body diagram
The mistake: adding an extra arrow labelled Fc next to tension, friction or gravity.
Why it is wrong: centripetal force is not a new force. It is the name for the net inward force, which real forces already supply. Adding it counts the same force twice.
How to spot it: every arrow on a free body diagram must name an object that exerts it ("rope on ball", "road on car"). "Centripetal" names no object, so it does not belong.
An "outward" (centrifugal) force
The mistake: "The passenger is thrown outward by a force pushing out of the curve."
Why it is wrong: no object pushes the passenger outward. The passenger's body tends to keep moving straight (inertia), and the car turns inward underneath. The door pushes the passenger inward.
How to spot it: in the ground frame, the net force in uniform circular motion always points toward the center. An outward arrow is a red flag.
"Constant speed means no acceleration"
The mistake: saying an object in uniform circular motion is in equilibrium.
Why it is wrong: acceleration is the change of the velocity vector. The direction changes, so a = v²/r, not zero, and the net force is not zero.
How to spot it: if your forces add to zero for an object moving in a circle, you have left something out or drawn something wrong.
Flying straight out after release
The mistake: thinking a released object moves directly away from the center.
Why it is wrong: after release no force acts sideways, so the object keeps its velocity at that instant, which was tangent to the circle.
How to spot it: draw the velocity arrow at the release point. The new path follows that arrow.
Wrong direction for the normal force at the top of a loop
The mistake: writing N − m g = m v²/r at the top, as if the track pushed up.
Why it is wrong: upside down at the top, the track is above the car and pushes it down, toward the center. So N + m g = m v²/r.
How to spot it: first mark where the center is. Forces toward the center are positive; forces away are negative.
Using diameter instead of radius, or mixing up period and frequency
The mistake: using the width of the circle as r, or writing v = 2πr × T.
Why it is wrong: r is from the center to the object; v = 2πr / T where T is seconds per revolution (frequency f = 1/T is revolutions per second).
How to spot it: check units: 2πr / T gives m/s. 2πr × T gives m·s, which is not a speed.